# Prove that L[f' ' ](s)$=$sL[f](s)

Can anyone prove this question ?

Let $f$:$\mathbb{R}$$→$$\mathbb{C}$ be continuous function such that $f$$(0) = 0 and that f' be a piecewise continuous function and absolutely integrable on \mathbb{R}. Prove that L[f'](s) = sL[f](s) where L[f] represents the Laplace transform of f - Did you try integration by parts? – mrf May 6 '13 at 21:40 what would be my u and v values, if I applied the integration by parts method – S F May 6 '13 at 21:41 Why don't you begin by writing down the definition of the Laplace transform? – wj32 May 6 '13 at 21:41 f(s)= L(ft)(s)$$ \int^inf_0 dx$ – S F May 6 '13 at 21:44
And what does the LHS of the identity you're trying to prove look like? – wj32 May 6 '13 at 21:47

$L\{f'(t)\}=\int_0^\infty e^{-st}f'(t)dt$

Integrating by parts we have,

$L\{f'(t)\}=e^{-st}f(t)|_0^\infty+s\int_0^\infty e^{-st}f(t)dt$

$L\{f'(t)\}=e^{-s(\infty)}f(\infty)-e^{-s(0)}f(0)+sL\{f(t)\}$

If $e^{-st}$ grows more rapidly than $f(t)$, we have $e^{-st}f(t)\to0$ when $t\to\infty$

$L\{f'(t)\}=sL\{f(t)\}-f(0)$

Since $f(0)=0$, this reduces to

$L\{f'(t)\}=sL\{f(t)\}$

-
thanks, do you know of any online resource that deals with this type of question ? – S F May 6 '13 at 21:54
I think Wiki is sufficient when studying basic Laplace theorem proofs. In the case of this question, wiki goes: en.wikipedia.org/wiki/… – Maazul May 6 '13 at 22:02